Question

For the reaction:
\[ Cl_2 + KOH \rightarrow KCl + KClO + H_2O \] 1 mole of Cl$_2$ is passed into 2 litre, 2 M KOH solution. Determine the molarity of Cl$^-$, ClO$^-$ and OH$^-$ respectively.

• A. 1 M, 0.5 M, 0.5 M
• B. 0.5 M, 0.5 M, 1 M
• C. 1 M, 1 M, 0.5 M
• D. 0.5 M, 1 M, 0.5 M

Hint:

Always check whether the reaction occurs in hot or cold alkali — products change accordingly.

Answer 1

The Correct Option is B

Step 1: Balanced reaction in cold KOH.
\[ Cl_2 + 2KOH \rightarrow KCl + KClO + H_2O \]
Step 2: Initial moles of reactants.
Moles of Cl$_2$ = 1
Moles of KOH = $2 \times 2 = 4$
Step 3: Stoichiometric consumption.
1 mole Cl$_2$ consumes 2 moles KOH
Remaining KOH = $4 - 2 = 2$ moles
Step 4: Calculation of molarity.
Moles of Cl$^-$ formed = 1
Moles of ClO$^-$ formed = 1
Volume = 2 L
\[ [\text{Cl}^-] = \frac{1}{2} = 0.5 \text{ M} \] \[ [\text{ClO}^-] = \frac{1}{2} = 0.5 \text{ M} \] \[ [\text{OH}^-] = \frac{2}{2} = 1 \text{ M} \]